Analysis Help - 1

IV. 1. Explain why it is that if a series converges (pointwise) on the closed interval [a, b] and converges uniformly on the open interval (a, b), then the series must converge uniformly on the closed interval [a, b]. (Note that to answer this question you must provide a proof that the series converges uniformly at the endpoints of the interval.)


IV. 2. Give an example of a function series that converges (pointwise) at every point of (a, b), each summand is continuous at every point in [a, b], but the series does not converge at every point of [a, b]. Formally prove that your function series satisfies all these conditions.

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Originally Posted by pila0688

IV. 1. Explain why it is that if a series converges (pointwise) on the closed interval [a, b] and converges uniformly on the open interval (a, b), then the series must converge uniformly on the closed interval [a, b]. (Note that to answer this question you must provide a proof that the series converges uniformly at the endpoints of the interval.)

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For the series $\displaystyle \{s_n(x)\}$ to converge uniformly to $\displaystyle s(x)$ on $\displaystyle (a,b)$ for all $\displaystyle \varepsilon>0$ there exists an $\displaystyle N_{\varepsilon}$ such that for all $\displaystyle n>N_{\varepsilon}$:

$\displaystyle |s_n(x)-s(x)|<\varepsilon$

and $\displaystyle N_{\varepsilon}$ is independednt of $\displaystyle x$.

As the series converges at $\displaystyle x=s$ and $\displaystyle x=b$, for all $\displaystyle \varepsilon>0$ there exist $\displaystyle N_{1,\varepsilon}$ and $\displaystyle N_{2,\varepsilon}$ such that for all $\displaystyle n>N_{1,\varepsilon}$:

$\displaystyle |s_n(a)-s(a)|<\varepsilon$

and for all $\displaystyle n>N_{2,\varepsilon}:$

$\displaystyle |s_n(b)-s(b)|<\varepsilon.$

Now for $\displaystyle \varepsilon>0$ let $\displaystyle M_{\varepsilon}=\max(N_{\varepsilon}$, $\displaystyle N_{1,\varepsilon},N_{2,\varepsilon}).$

Now for all $\displaystyle x$ in $\displaystyle [a,b]$ we have for all $\displaystyle n>M_{\varepsilon}:$

$\displaystyle |s_n(x)-s(x)|<\varepsilon$

which proves that $\displaystyle \{ s_n(x) \}$ is uniformly convergent on $\displaystyle [a,b]$

CB